The explanation for the solution is pretty half-arsed. There is no way to do it except by trial and error, and his suggestion that you'd guess the correct answer quickly is a bit tenuous. That said it is exactly what I did.

Some fractions are not recurring. And he's implying that there are more recurring fractions than non-recurring ones.

My own line of thought is that the number of non-recurring fractions is exactly equal to the number of recurring ones. I think both sets are countably infinite and that as a result there is a one to one mapping between the two sets.

The set of all fractions (aka the set of all rational numbers, aka the set of all decimals that either terminate or recur) is known to be countably infinite, ie each element of the set can be mapped with a bijection to the set of natural numbers.

I think that the set of terminating decimals is also countably infinite. It's certainly infinite (0.1, 0.01, 0.001 ...) and since it is a subset of the rational numbers I don't believe it can be uncountably infinite (citation needed). Therefore it is countably infinite. And hence the "number" of fractions with a recurring decimal representation is equal to the "number" of fractions with a terminating decimal representation. And so his statement that "almost all" fractions have a recurring decimal representation is false.

this falls apart at part a, because you're suggesting multiple decimal representations for the same actual number.

If that's "allowed" then his statement is just trivially obvious, because every terminating decimal would have its own infinite set of non-terminating equivalents.

But it isn't allowed. The decimal representation of a number is the one that is expressed in the simplest terms. So no trailing zeros and no 9 recurrings. That way each number has one unique decimal representation.

"Some real numbers have two infinite decimal representations. For example, the number 1 may be equally represented by 1.000... as by 0.999... (where the infinite sequences of digits 0 and 9, respectively, are represented by "..."). Conventionally, the version with zero digits is preferred; by omitting the infinite sequence of zero digits, removing any final zero digits and a possible final decimal point, a normalized finite decimal representation is obtained."

Basically, the only fractions that DON'T recur in base 10 are ones which, as you noted above, can be written in the form "a/(10^n)".

But infinite-set cardinality is counter-intuitive: The sole reason for the cardinality of the real line being greater than that of integers is down to transcendental numbers, which are difficult to calculate in spite of being numerically abundant.

Thus, INTUITIVELY you would say that there are lots more fractions that recur than ones that of the "a/(10^n)" form. When you're doing year 10 level maths, you're unlikely to be particularly familiar with number theory and the cardinality of infinite sets.

So yeah, whilst I'd agree that fundamentally he's incorrect, for the purposes of what you'd be teaching year 10s he's right enough.

so that's my little side-argument dealt with. Both sets are N0 cardinality, and the concept of "more than" doesn't really make sense in the context he's using it in.

But taking things a bit more generally and less rigorously I'm still intrigued by his implication.

I would argue that the most commonly found fractions (by a year 10 or anyone else) are half, third, ..., tenth. Of which only four out of nine are recurring decimals. So even on that level I think he's talking out of his arse.

And this is a reader in maths we're talking about. He really shouldn't be saying stuff that's wrong at all, even if it is intuitive.

you just have to know your squares of numbers and then have to exeriment with them a bit...unless you can 'see' I didn't see it immediately, I just experimented with squares ending in 1 so it took me 3 trials to get the answer